devarshi-dt-logo

Question:

ABC is a triangular park with AB=AC=100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are cos⁻¹(3/2) and cosec⁻¹(2/2) respectively, then the height of the tower (in metres is):

105

100√3

20

25

Solution:

Correct option is C. 20
cot α = 3/2
cosec β = 2/2
So, x/h = 3/2.. (i)
and h/√100² - x² = 1/2.. (ii)
So, from (i) and (ii), h/√100² - (3h/2)² = 1/2
25h² = 100 × 100 ⇒ h = 20